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Consider the following integral: $$J(u)=\int_{0}^{u^2}\int_{0}^{u}\frac{1}{x^3+y+3x^2+5x+3}dxdy$$ My question: I want to separate it into two parts: $$J(u)=u^aI(u)$$ where $a\geq 0$ and $I(u)$ is "smaller" than polynomials, that is, $I(u)$ satisfies the followings:

(1) $\liminf\limits_{u\rightarrow+\infty}I(u)>0$. (2) for any $p>0$, $\lim\limits_{u\rightarrow+\infty}\frac{I(u)}{u^p}=0$

For example, $I(u)=\log(u)$, or $\arctan u$ etc.

My attempt: I first guess $a=0$, that is, $J(u)$ itself satisfies $I(u)$. But I can't show that property (2) above. Then I guess $a=1$, that is, $J(u)=uI(u)$, I can show that property (2) by change the variables $(\frac{x}{u},\frac{y}{u^2})=(w,s)$ and using Dominated convergence theorem. But this case I can't show the property (1)... And now I use computer to calculate, I guess $a$ may be between $(0,1)$.

Some tips:

(i) if you change the variables, $(\frac{x}{u},\frac{y}{u^2})=(w,s)$ may be better. this is called dilation and it is a kind of "homogeneous"

(ii) If I change $x^3$ into $xy$ in $J(u)$, I can show that actually $a=0$, by Dominated convergence theorem.

Background: Given a dilation in $R^n$: $$D_t(x)=(t^{c_1}x_1,...,t^{c_n}x_n)$$ then consider $P(x,r)=r^nf_n(x)+...+r^Qf_Q(x)$, where $Q=m_1+...+m_n$, $f_k(x)$ satisfies $f_k(D_t(x))=t^{Q-k}f_k(x)$, and $f(x)>0$ are the combinations of positive monomials ($x_i>0$). The origin type of the denominator of the $J(u)$ is $x^3r^3+(y+3x^2)r^4+5xr^5+3r^6$. And the dilation is $D_t(x)=(t^{1}x,t^2y,t^{3}z)$. I just take $1/r=u$ and do some simplification.

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  • $\begingroup$ One direct simplification is to change the order of integration, to get $$ \begin{split} J(u) &= \int_0^{u^2}\int_0^u\frac{dxdy}{x^3+y+3x^2+5x+3} \\ &= \int_0^u \left[ \int_0^{u^2}\frac{dy}{x^3+y+3x^2+5x+3} \right] dx \\ &= \int_0^u \ln \left(\frac{x^3+3x^2+5x+3} {u^2+x^3+3x^2+5x+3} \right) dx \\ &= \int_0^u \ln \left(1 - \frac{u^2} {u^2+x^3+3x^2+5x+3} \right) dx \\ \end{split} $$ $\endgroup$
    – gt6989b
    Jun 14, 2020 at 4:49
  • $\begingroup$ but, maybe you calculate wrong? the integral should be positive but yours is nagetive... and if you calculate in the right way you will find that's not a good way to due with the integral $\endgroup$
    – Houa
    Jun 14, 2020 at 5:05
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    $\begingroup$ We have $$J(u) = \int_0^u \ln \left( 1 + \frac {u^2} {P(x)} \right) dx, \\ J'(u) = 2 u \int_0^u \frac {dx} {P(x) + u^2} + \ln \left( 1 + \frac {u^2} {P(u)} \right).$$ The integral in $J'(u)$ (which is similar to this integral) can be brute-forced by writing out an antiderivative in terms of a sum over the roots of $P(x) + u^2$ and taking an asymptotic expansion of the roots for large $u$. This gives $$J'(u) \sim \frac {4 \pi} {3 \sqrt 3 \hspace {1.5px} u^{1/3}},$$ therefore $$J(u) \sim \frac {2 \pi u^{2/3}} {\sqrt 3}$$ by l'Hopital's rule. $\endgroup$
    – Maxim
    Jul 26, 2020 at 20:43
  • $\begingroup$ @Maxim Good idea by l'Hopital's rule! By the way, do you have any reference to due with such integrals? This is just a concrete example. The more general type is like the description in the "background" in the body above.($P(x,r)=....$) $\endgroup$
    – Houa
    Jul 27, 2020 at 1:00

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